Tag: Improper Integrals

Test the improper integral ∫ e^-x/2 for convergence

by RoRi

March 31, 2016

Test the following improper integral for convergence:

$\int_0^{\infty} \frac{1}{\sqrt{e^x}} \, dx.$

The integral converges.

Proof. We compute directly,

$\begin{align*} \int_0^{\infty} \frac{1}{\sqrt{e^x}} \, dx &= \int_0^{\infty} e^{-\frac{x}{2}} \, dx \\[9pt] &= \lim_{b \to +\infty} \left( \int_0^b e^{-\frac{x}{2}} \, dx \right) \\[9pt] &= \lim_{b \to +\infty} \left( -2 e^{-\frac{x}{2}} \Bigr \rvert_0^b \right) \\[9pt] &= \lim_{b \to +\infty} \left( -2 e^{-\frac{b}{2}} + 2 \right) \\[9pt] &= 2. \qquad \blacksquare \end{align*}$

Test the improper integral ∫ 1 / (x³ + 1)^1/2 for convergence

by RoRi

March 31, 2016

Test the following improper integral for convergence:

$\int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx.$

The integral converges.

Proof. We know the integral $\int_0^{\infty} \frac{1}{x^{\frac{3}{2}}} \, dx$ converges (example #1 on page 417 of Apostol). Applying the limit comparison test (by the note to Theorem 10.25 on page 418, which says that if $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$ then the convergence of $\int g(x)$ implies the convergence of $\int f(x)$ .) we have

$\begin{align*} \lim_{x \to +\infty} \frac{f(x)}{g(x)} &= \lim_{x \to +\infty} \frac{ \frac{1}{\sqrt{x^3+1}}}{\frac{1}{x}} \\[9pt] &= \lim_{x \to +\infty} \frac{x}{\sqrt{x^3+1}} \\[9pt] &= \lim_{x \to +\infty} \frac{1}{x^{\frac{1}{2}} \sqrt{1 + \frac{1}{x}}} \\[9pt] &= 0. \end{align*}$

Since we know $\int_0^{\infty} \frac{1}{x^{\frac{3}{2}}} \, dx$ converges the theorem establishes the convergence of

$\int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx. \qquad \blacksquare$

Test the improper integral ∫ e^-x² for convergence

by RoRi

March 31, 2016

Test the following improper integral for convergence:

$\int_{-\infty}^{\infty} e^{-x^2} \, dx.$

The integral converges.

Proof. We have

$\begin{align*} \int_{-\infty}^{\infty} e^{-x^2} \, dx &= \int_{-\infty}^0 e^{-x^2} \, dx + \int_0^{\infty} e^{-x^2} \, dx \\[9pt] &= 2 \int_0^{\infty} e^{-x^2} \, dx &(e^{-x^2} \text{ is an even function})\\[9pt] &= 2 \left( \int_0^1 e^{-x^2} \, dx + \int_1^{\infty} e^{-x^2} \, dx \right). \end{align*}$

The first integral converges since it is a proper integral. For the second integral, since $x \geq 1$ we have $e^{-x^2} \leq e^{-x}$ . But, we know $\int_1^{\infty} e^{-x} \, dx$ converges (by Example #4 on page 418 of Apostol with $s = 0$ ). Hence, we have established the convergence of

$\int_{-\infty}^{\infty} e^{-x^2} \, dx. \qquad \blacksquare$

Test the improper integral ∫ x / (x⁴ + 1)^1/2 for convergence

by RoRi

March 31, 2016

Test the following improper integral for convergence:

$\int_0^{\infty} \frac{x}{\sqrt{x^4+1}} \, dx.$

This integral diverges.

Proof. By example 1 (page 417) of Apostol, we know the integral $\int_1^{\infty} \frac{1}{x} \, dx$ diverges. We then apply the limit comparison test (Theorem 10.25 on page 418),

$\begin{align*} \lim_{x \to +\infty} \frac{f(x)}{g(x)} &= \lim_{x \to +\infty} \frac{\frac{x}{\sqrt{x^4+1}}}{\frac{1}{x}} \\[9pt] &= \lim_{x \to +\infty} \frac{x^2}{\sqrt{x^4+1}} \\[9pt] &= \lim_{x \to +\infty} \frac{1}{\sqrt{1+\frac{1}{x^4}}} \\[9pt] &= 1. \end{align*}$

Thus, by the limit comparison test, we have established the divergence of

$\int_0^{\infty} \frac{x}{\sqrt{x^4+1}} \, dx. \qquad \blacksquare$